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I have seen this in a book I am reading it seems that this is straight forward, but unfortunately my algebraic skills not good enough.

$A$ is symmetric with zeros on its diagonal, we need to prove that there exist matrix $U$ s.t $$ A+2 I=U U^{T} $$

misha312
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A counterexample for real matrices is $A=\begin{pmatrix} 0 & 3 \\ 3 & 0 \end{pmatrix}$. Note that, if $A+2I=UU^t$ for some matrix $U$, then for any real vector $v$ we have $$v^t (A+2I)v=v^tUU^t v=(U^t v)^t (U^tv)=\|U^t v\|^2$$ which cannot be negative if $U$ (and therefore $U^t v$) is real. But if $v=(1,-1)^t$ then $$v^\top (A+2I)v = \begin{pmatrix} 1 & -1 \end{pmatrix} \begin{pmatrix} 2 & 3\\3 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix} =\begin{pmatrix} -1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix}=-2. $$ Hence no real matrix $U$ exists. Matters are different if we allow complex-valued $U$. For instance, Mathematica's FindInstance command generates the example $U=\begin{pmatrix} 0 & -\sqrt{2} \\ -i \sqrt{5/2} & -3/\sqrt{2}\end{pmatrix}.$

Semiclassical
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    If $U$ is allowed to be complex, then I would suspect that the transpose should really be a Hermitian conjugate. But then the conjecture is definitely false. Each row of $U$ would have norm $\sqrt{2}$, so if any element of $A$ had a greater magnitude, then $A$ would be a counterexample. So there doesn’t seem to be any way to save this without imposing additional conditions on $A$. – sasquires Feb 06 '21 at 21:32
  • @sasquires Agreed. One needs $A+2I$ to be PSD to forbid the kind of counterexample I give above. So if none of the eigenvalues of $A$ are smaller than $-2$, then it should be fine. – Semiclassical Feb 07 '21 at 01:13